Which of the following numbers is a multiple of 3? ${53,57,97,106,109}$
Explanation: The multiples of $3$ are $3$ $6$ $9$ $12$ ..... In general, any number that leaves no remainder when divided by $3$ is considered a multiple of $3$ We can start by dividing each of our answer choices by $3$ $53 \div 3 = 17\text{ R }2$ $57 \div 3 = 19$ $97 \div 3 = 32\text{ R }1$ $106 \div 3 = 35\text{ R }1$ $109 \div 3 = 36\text{ R }1$ The only answer choice that leaves no remainder after the division is $57$ $ 19$ $3$ $57$ We can check our answer by looking at the prime factorization of both numbers. Notice that the prime factors of $3$ are contained within the prime factors of $57$ $57 = 3\times19 3 = 3$ Therefore the only multiple of $3$ out of our choices is $57$. We can say that $57$ is divisible by $3$.